Calculus

Differentiability

Differentiability: Level 4 Challenges

         

λ=(f(5102)f(2015)f(c))(f2(2015)+f2(5102)+f(2015)f(5102)f2(c))\lambda=\left( \frac{f(5102)-f(2015)}{f'(c)} \right) \left( \frac{f^2(2015)+f^2(5102)+f(2015)f(5102)}{f^2(c)} \right)

Let f:[2015,5102][0,)f:[2015,5102] \rightarrow [0,\infty)be any continuous and differentiable function. Find the value of λ\lambda, such that there exists some c[2015,5102]c\in [2015,5102] which satisfies the equation above.

f(x)=i=11+2cos(2x3i)3 f(x) = \prod_{i=1}^{\infty} \frac{1+ 2\cos\left( \frac{2x}{3^i} \right)}{3}

Let f(x)f(x) denote an function as described above. The number of points where xf(x)+x21 |xf(x)| + | |x-2|-1| is non-differentiable in x(0,3π) x \in (0,3\pi) is kk, find k2k^2.

(α1,β1), (α2,β2),  , (αn,βn)(\alpha_1, \beta_1), \ (\alpha_2, \beta_2), \ \cdots \ , \ (\alpha_n, \beta_n) are the points on the curve S : y=sin(x+y)  x[4π, 4π]S \ : \ y=\sin (x+y) \ \ \forall x \in [-4\pi, \ 4\pi]
such that the tangents at these points to SS are perpendicular to the line l : (21)x+y=0.l \ : \ (\sqrt {2}-1)x+y=0.

Find the value of k=1nαkβk.\left\lfloor \displaystyle\sum_{k=1}^{n} |\alpha_k|-|\beta_k|\right\rfloor.

Details and Assumptions

  • \lfloor{\cdots}\rfloor denotes the floor function (greatest integer function).

  • Even though in the problem, a plural form of points is given, there may only be one such point that exists.

What is the largest possible number of integers nn such that for some fixed polynomial f(x)f(x) of degree 33 with integer coefficients, f(n)=n1000f(n)=n^{1000}?

Let f(x)f(x) be a non-constant thrice differentiable function defined on real numbers such that f(x)=f(6x)f(x)=f(6-x) and f(0)=0=f(2)=f(5)f'(0)=0=f'(2)=f'(5). Find the minimum number of values of p[0,6]p \in [0,6] which satisfy the equation (f(p))2+f(p)f(p)=0(f''(p))^2+f'(p)f'''(p)=0 Details and Assumptions:

  • f(p)=(df(x)dx)x=pf'(p)=\left( \frac{df(x)}{dx} \right)_{x=p}

  • f(p)=(d2f(x)dx2)x=pf''(p)=\left( \frac{d^2f(x)}{dx^2} \right)_{x=p}

  • f(p)=(d3f(x)dx3)x=pf'''(p)=\left( \frac{d^3f(x)}{dx^3} \right)_{x=p}

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